In Fig., a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. the lower compartment of the container is filled with 2 moles of an ideal monoatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal diatomic gas at 400 K. the heat capacities per mole of an ideal monoatomic gas are `C_(upsilon) = (3)/(2) R and C_(P) = (5)/(2) R`, and those for an ideal diatomic gas are `C_(upsilone) = (5)/(2) R and C_(P) = (7)/(2) R.`
Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. the total work done by the gases till the time they achieve equilibrium will be

Options

A. 250 R
B. 200 R
C. 100 R
D. `-100 R`

Answer

D

Explanation

In this case, pressure of each gas is contant.
Let T be the final temp. of each gas in equilibrium
Heat gained by diatomic gas
= Heat lost by monoatomic gas
`n_(1) xx C_(P_1)(T-400) = n_(2) xx C_(P_2)(700-T)`
`2 xx (7)/(2) R(T-400) = 2 xx (5)/(2)R(700-T)`
`7 T-2800 = 3500 - 5 T`
`12 T = 3500 + 2800 = 6300`
` T = (6300)/(12) = 525 K`
Change in internal energy of the system
`(delta)U = n_(1) C_(upsilon_1) (T - 400) + N_(2) C_(upsilon_2) (T - 700)`
` = 2 xx (5)/(2)R (525 - 400) + 2xx(3)/(2)R (525 - 700)`
`625 R - 525 R = 100 R`
As no heat is supplied from outside , `(delta)Q = 0`
`(delta)W = -(delta)U = -100 R`